Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k + 6}{k^2 + 15k + 50} \div \dfrac{-6k - 36}{-9k^2 - 90k} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k + 6}{k^2 + 15k + 50} \times \dfrac{-9k^2 - 90k}{-6k - 36} $ First factor the quadratic. $q = \dfrac{k + 6}{(k + 10)(k + 5)} \times \dfrac{-9k^2 - 90k}{-6k - 36} $ Then factor out any other terms. $q = \dfrac{k + 6}{(k + 10)(k + 5)} \times \dfrac{-9k(k + 10)}{-6(k + 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k + 6) \times -9k(k + 10) } { (k + 10)(k + 5) \times -6(k + 6) } $ $q = \dfrac{ -9k(k + 6)(k + 10)}{ -6(k + 10)(k + 5)(k + 6)} $ Notice that $(k + 6)$ and $(k + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -9k(k + 6)\cancel{(k + 10)}}{ -6\cancel{(k + 10)}(k + 5)(k + 6)} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $q = \dfrac{ -9k\cancel{(k + 6)}\cancel{(k + 10)}}{ -6\cancel{(k + 10)}(k + 5)\cancel{(k + 6)}} $ We are dividing by $k + 6$ , so $k + 6 \neq 0$ Therefore, $k \neq -6$ $q = \dfrac{-9k}{-6(k + 5)} $ $q = \dfrac{3k}{2(k + 5)} ; \space k \neq -10 ; \space k \neq -6 $